噪声运算符的扩展

16

在我当前正在解决的一个问题中，噪声运算符的扩展自然而然地出现了，我很好奇是否已经进行了先前的工作。首先让我修改的基本噪声操作 $T_{\varepsilon}$ 真实值的布尔函数。给定的函数 $f: \{0,1\}^n \to \mathbb{R}$ 和 $\varepsilon$ ， $p$ ST $0 \leq \varepsilon \leq 1$ ， $\varepsilon = 1 - 2p$ ，我们定义 $T_{\varepsilon} \to \mathbb{R}$ 为 $T_{\varepsilon} f(x) = E_{y \sim \mu_p} [f(x+y)]$

$\mu_p$ 是在分配 $y$ 通过设置的每个位而获得 $n$ 位矢量为 $1$ 独立地以概率 $p$ 和 $0$ 否则。同样，我们可以将这个过程视为以独立的概率翻转每一位。现在，这个噪声运营商具有许多有用的特性，包括作为乘法和具有很好的特征向量（ $x$ $p$ $T_{\varepsilon_1} T_{\varepsilon_2} = T_{\varepsilon_1 \varepsilon_2}$ 其中属于奇偶基础）。 $T_{\varepsilon}(\chi_S) = \varepsilon^{|S|} \chi_S$ $\chi_S$

现在让我定义我延伸，这是我表示为。由下式给出。但是这里我们的分布 $T_\varepsilon$ $R_{(p_1,p_2)}$ $R_{(p_1,p_2)} \to \mathbb{R}$ $R_{(p_1,p_2)} f(x) = E_{y \sim \mu_{p,x}} [f(x+y)]$ 是这样的，我们的翻转的比特到的概率为和的比特到的概率为。（是现在显然分布依赖于其中函数进行评估，并且如果，然后降低到“常规”噪声操作者）。 $\mu_{p,x}$ $1$ $x$ $0$ $p_1$ $0$ $x$ $1$ $p_2$ $\mu_{p,x}$ $x$ $p_1 = p_2$ $R_{(p_1,p_2)}$

我想知道，这个运算符已经在文献中被很好地研究过？还是它的基本特性显而易见？我只是从布尔分析开始，所以对比我更熟悉该理论的人可能很简单。我特别感兴趣的是，特征向量和特征值是否具有很好的特征，或者是否存在任何可乘性。 $R_{(p_1,p_2)}$

boolean-functions fourier-analysis

— 阿米尔
source

14

我将回答问题的第二部分。

I.特征值和特征函数

我们首先考虑一维情况。很容易检查算子具有两个本征函数：和，特征值和 $n=1$ $R_{p_1,p_2}$ $1$

ξ (x) = (p_{1} + p_{2}) x - p_{1} = {\begin{cases} - p_{1}, & if x = 0, \\ p_{2}, & if x = 1. \end{cases}

$\xi(x) = (p_1+p_2)x - p_1 = \begin{cases} -p_1, &\text{ if } x =0,\\ p_2, &\text{ if } x =1. \end{cases}$

1

$1$

。

1 - p_{1} - p_{2}

$1-p_1 - p_2$

现在考虑一般情况。对于，让。观察到，是本征函数。实际上，由于所有变量是独立的，因此我们有 $S\subset \{1,\dots,n\}$ $\xi_S(x) = \prod_{i\in S} \xi(x_i)$ $\xi_S$ $R_{p_1,p_2}$ $x_i$

\begin{aligned} R_{p_{1}, p_{2}} (ξ (x)) & = R_{p_{1}, p_{2}} (\prod_{i \in S} ξ (x_{i})) = \prod_{i \in S} R_{p_{1}, p_{2}} (ξ (x_{i})) \\ = \prod_{i \in S} ((1 - p_{1} - p_{2}) ξ (x_{i})) = (1 - p_{1} - p_{2})^{| S |} ξ_{S} (x) . \end{aligned}

$\begin{align*} R_{p_1,p_2}(\xi(x)) &= R_{p_1,p_2}\left(\prod_{i\in S} \xi(x_i)\right) = \prod_{i\in S} R_{p_1,p_2}(\xi(x_i)) \\ &= \prod_{i\in S}\left((1-p_1 - p_2)\xi(x_i)\right) = (1-p_1 - p_2)^{|S|} \xi_S(x). \end{align*}$

我们得到是的本征函数与特征值对于每个。由于函数跨度的整个空间， $\xi_S(x)$ $R_{p_1,p_2}$ $(1-p_1-p_2)^{|S|}$ $S\subset \{1,\dots,n\}$ $\xi_S(x)$ $R_{p_1,p_2}$ 没有其他的本征函数（不属于的线性组合）。 $\xi_S(x)$

二。乘性

一般而言，“乘法属性”不保持自的eigenbasis 取决于和。但是，我们有其中 $R_{p_1,p_2}$ $R_{p_1,p_2}$ $p_1$ $p_2$

R_{p_{1}, p_{2}}^{2} = R_{p_{1}^{'}, p_{2}^{'}},

$R_{p_1,p_2}^2 = R_{p_1',p_2'},$

和

。为了验证，首先注意

和

具有相同的组的本征函数

。我们有，

p_{1}^{'} = 2 p_{1} - (p_{1} + p_{2}) p_{1}

$p_1' = 2p_1- (p_1+p_2)p_1$

p_{2}^{'} = 2 p_{2} - (p_{1} + p_{2}) p_{2}

$p_2' = 2p_2- (p_1+p_2)p_2$

R_{p_{1}, p_{2}}

$R_{p_1,p_2}$

R_{p_{1}^{'}, p_{2}^{'}}

$R_{p_1',p_2'}$

{ξ_{S}}

$\{\xi_S\}$

R_{p_{1}, p_{2}}^{2} (ξ_{S}) = (1 - p_{1} - p_{2})^{2 | S |} ξ_{S} = (1 - p_{1}^{'} - p_{2}^{'})^{| S |} ξ_{S} = R_{p_{1}^{'}, p_{2}^{'}} (ξ_{S})

$R_{p_1,p_2}^2(\xi_S) = (1-p_1-p_2)^{2|S|} \xi_S=(1-p_1'-p_2')^{|S|}\xi_S = R_{p_1',p_2'} (\xi_S)$

\begin{aligned} 1 - p_{1}^{'} - p_{2}^{'} & = 1 - p_{1} \cdot (2 - (p_{1} + p_{2})) - p_{2} \cdot (2 - (p_{1} + p_{2})) \\ = 1 - (p_{1} + p + 2) (2 - (p_{1} + p_{2})) \\ = 1 - 2 (p_{1} + p_{2}) + (p_{1} + p_{2})^{2} = (1 - p_{1} - p_{2})^{2} . \end{aligned}

$\begin{align*} 1-p_1'-p_2' &= 1 - p_1\cdot (2- (p_1+p_2)) - p_2\cdot (2- (p_1+p_2)) \\ &= 1 - (p_1+p+2) (2- (p_1+p_2)) \\ &= 1 - 2(p_1 +p_2) + (p_1 + p_2)^2 = (1-p_1-p_2)^2. \end{align*}$

III. Relation to the Bonami—Beckner operator

Let us think of functions from $\{0,1\}^n$ to ${\mathbb R}$ as polylinear polynomials. Let $\delta = \frac{1}{2}\cdot \frac{p_1-p_2}{p_1+p_2}$ . Consider the operator

A_{δ} (f) = f (x_{1} + δ, \dots, x_{n} + δ) .

$A_{\delta}(f) = f(x_1 + \delta, \dots, x_n + \delta).$ It maps every multilinear polynomial

f

$f$ to a multilinear polynomial

A [f]

$A[f]$ . We have,

R_{p_{1}, p_{2}} (f) = A_{δ}^{- 1} T_{ε} A_{δ} (f),

$R_{p_1,p_2}(f) = A_{\delta}^{-1} T_{\varepsilon}A_{\delta}(f),$ where

ε = 1 - p_{1} - p_{2}

$\varepsilon = 1 - p_1 - p_2$ . Note that parts I and II follow from this formula and properties of the Bonami—Beckner operator.

— Yury
source

Yury, thank you for the answer! That's a good starting point for me to work with; I should now be able to work out if there are analogues of the hyper contractive inequality. Will post back here if I get any more interesting analysis.

— Amir

This is very long after the fact, but I am curious how you derived the third part and the relation to the Becker Bonami operator?

— 阿米尔（Amir）2013年

(a) It is sufficient to check the identity for

f = 1

$f=1$ and

f = x_{i}

$f=x_i$ . If it holds for

1

$1$ and

x_{i}

$x_i$ , then it's easy to see that it holds for all characters. By linearity, it holds for all functions. (b) Alternatively, from I,

T_{ε}

$T_\varepsilon$ and

R_{p_{1}, p_{2}}

$R_{p_1,p_2}$ have the same set of eigenvalues; eigenvector

\prod_{i \in S} x_{i}

$\prod_{i\in S} x_i$ of

T

$T$ “corresponds” to eigenvector

\prod_{i \in S} ξ (x_{i})

$\prod_{i\in S} \xi(x_i)$ of

R

$R$ . Thus

R (f) = A^{- 1} T A (f)

$R(f) = A^{-1} T A(f)$ where A is a linear map that maps

ξ (x)

$\xi(x)$ to

x

$x$ .

— Yury

3

We were eventually able to analyze hypercontractive properties of $R_{p_1, p_2}$ (http://arxiv.org/abs/1404.1191), building off of the main Fourier analysis of $R_{p,0}$ by Ahlberg, Broman, Griffiths and Morris (http://arxiv.org/abs/1108.0310).

To summarize, the effect of a biased operator $R_{p,0}$ on a function $f$ can be analyzed as a symmetric noise operator in a biased measure space. This gives a weak form of hypercontractivity, which depends on how the $\ell_2$ norm of $f$ varies when switching to a choice of biased measure $\mu$ dependent on $p$ .

— Amir
source

You might want to 'accept' this answer so that the question doesn't keep popping up (disclaimer: I am an author on the linked paper)

— Suresh Venkat